which triangle do you mean? , where is it?
tan [ theta(S1) ] is (σx- σy) / 2 τxy , where ##A=\tau_{xy}## and ##O=\frac{1}{2}(\sigma_x-\sigma_y)##

Simon means the grey triangle on the right, and I agree with his equation.
The tangent will be negative, and it will be vertical displacement divided by horizontal displacement, so ##\frac{2\tau}{\Delta \sigma}##.
You can check this by thinking about the coordinates of the top corner of the left-hand grey triangle in polar: ##y=r\sin(\theta)##, ##x=r\cos(\theta)##.

Simon means the grey triangle on the right, and I agree with his equation.
The tangent will be negative, and it will be vertical displacement divided by horizontal displacement, so ##\frac{2\tau}{\Delta \sigma}##.
You can check this by thinking about the coordinates of the top corner of the left-hand grey triangle in polar: ##y=r\sin(\theta)##, ##x=r\cos(\theta)##.

so , the diagram is wrong ?
because it's tan [ theta(S1) ] is (σx- σy) / 2 τxy , where ##A=\tau_{xy}## and ##O=\frac{1}{2}(\sigma_x-\sigma_y)##
so , the triangle should be drawn like this rather the grey triangles given by the author , am i right?

Attached Files:

so , the diagram is wrong ?
because it's tan [ theta(S1) ] is (σx- σy) / 2 τxy , where ##A=\tau_{xy}## and ##O=\frac{1}{2}(\sigma_x-\sigma_y)##
so , the triangle should be drawn like this rather the grey triangles given by the author , am i right?

Merely considering a different triangle is not going to change the relationship.
To get the tangent to be sigma/tau you need theta to be measured from a tau axis, not from a sigma axis.

The only difference I can see between your original diagram and the one in post #5 is that two triangles have been coloured in red. Both diagrams show the theta angles by means of black curved arrows starting at the positive sigma (x) axis.

The only difference I can see between your original diagram and the one in post #5 is that two triangles have been coloured in red. Both diagrams show the theta angles by means of black curved arrows starting at the positive sigma (x) axis.

sorry , i forgot to label the angle in the new triangle that i drew , here's what i mean ...
According to the notes , tan( 2 θs1) = -(σx -σy) / 2τxy , are they correct now ?

how to make tan (2 θs1) and tan (2 θs2 ) positive ?
From the figure , we can see that tan( 2 θs1) and tan (2 θs2 ) = - (σx -σy) / 2τxy , both are negative ....

The diagram labels the distance from the vertical axis to the left hand dashed line as - (σx -σy), but it is not. As a distance it is + (σx -σy). The x coordinate of that dashed line is - (σx -σy), but that is another matter.
With standard Cartesian and polar coordinates, it is true that a point with coordinates (x, y) is at (r, θ), where y/x=tan(θ). But since we have had to switch theta to be measured from the positive vertical axis, we now have instead that x/y=-tan(θ).
(This is connected with the fact that the product of the gradients of two lines at right angles is -1.)

The diagram labels the distance from the vertical axis to the left hand dashed line as - (σx -σy), but it is not. As a distance it is + (σx -σy). The x coordinate of that dashed line is - (σx -σy), but that is another matter.
With standard Cartesian and polar coordinates, it is true that a point with coordinates (x, y) is at (r, θ), where y/x=tan(θ). But since we have had to switch theta to be measured from the positive vertical axis, we now have instead that x/y=-tan(θ).
(This is connected with the fact that the product of the gradients of two lines at right angles is -1.)